Past exam of the options II course of the University of Oxford / 2010 / F / F2 / c

Work on the positive orthant, where all real powers in $g$ are defined. To justify a global maximum, enlarge the constraint set to the compact simplexIt is closed and bounded since $x_i\le 1/p_i$. The product $g$ extends continuously to $K$, is zero at every boundary point where a coordinate vanishes, and equals $1$ at $(1,\dots ,1)$. Hence a maximum exists and lies in the positive interior. For $n=1$ the constraint gives $x_1=1$ directly.

For $n\ge 2$, at an interior maximum the Lagrange multiplier equations areAll $p_i$ and $g$ are positive, so all $x_i$ are equal. The constraint and ${\sum }_i{p}_i=1$ make their common value $1$. Thus the only possible maximizer has value $1$, provingwith equality exactly when every $x_i=1$. The source writes $g$ on all of ${\mathbb{R}}^n$; for general noninteger exponents its relevant real-valued domain is the positive orthant used here.

Let $S={\sum }_i{p}_i{a}_i>0$ and set $x_i=a_i/S$. Then $x_i>0$ and ${\sum }_i{p}_i{x}_i=1$, so part (1) givesTherefore the weighted arithmetic-geometric mean inequality isEquality holds exactly when all $a_i$ are equal, by the equality case in part (1).